CLASS 9 MATH CHAPTER 6 LINES AND ANGLES KAI QUESTION HAI PLEASE TELL KAISE KREGY??????????
Answers
Answer:
3. x = 50°
∠AOD = ( x + 10° ) = 50° + 10° = 60°
∠COD = x = 50°
∠BOC = x + 20° = 50° + 20° = 70°
5. a = 130°
b = 50°
10. x = 45°
9. x = 55°
8. x = 30°
Step-by-step explanation:
3. Given: ∠AOD = ( x + 10° ), ∠COD = x and ∠BOC = x + 20°.
We know that the sum of all the angles in a straight line is 180°
∠AOD + ∠COD + ∠BOC = 180°
( x + 10° ) + ( x ) + ( x + 20° ) = 180°
3x + 30° = 180°
3x = 180° - 30°
3x = 150°
x = 150° / 3
x = 50°
∠AOD = ( x + 10° ) = 50° + 10° = 60°
∠COD = x = 50°
∠BOC = x + 20° = 50° + 20° = 70°
5. Given: ∠BOC = b and ∠AOC = a form linear pair & a - 2b = 30°.
To find: a and b
∠AOC + ∠BOC = 180° ——————— [ Linear Pair ]
a + b = 180° ————— Equation (i)
Equation given :
a - 2b = 30° ————— Equation (ii)
Equation (ii) - Equation (i) gives
-3b = 30° - 180°
3b = 150°
b = 150° / 3
b = 50°
Substituting b = 50° in Equation (i)
a + 50° = 180°
a = 180° - 50°
a = 130°
∴ a = 130° and b = 50°
10. Given: POS is a line. ∠POQ = 60°, ∠QOR = 4x and ∠ROS = 40°
We know that the sum of all the angles in a straight line is 180°
∠POQ + ∠QOR + ∠ROS = 180°
60° + 4x + 40° = 180°
100° + 4x = 180°
4x = 180° - 100°
4x = 80°
x = 180° / 4
∴ x = 45°
9. Given: AOC is a line. ∠AOB = 70° and ∠BOC = 2x.
We know that the sum of all the angles in a straight line is 180°
∠AOB + ∠BOC = 180°
70° + 2x = 180°
2x = 180° - 70°
2x = 110°
x = 110° / 2
∴ x = 55°
8. In the given figure
3x + 3x + x + 150° = 360°
7x = 360° - 150°
7x = 210°
x = 210° / 7
x = 30°